10. Force Due to Liquid Pressure
by M. Bourne
The force F on an area A at a depth y in a liquid of density w is given by
F = wyA
The force will increase if the density increases, or if the depth increases or if the area increases.
So if we have an unevenly shaped plate submerged vertically in a liquid, the force on it will increase with depth. Also, if the shape of the plate changes as we go deeper, we have to allow for this.
So we have:
Now, the total force on the plate is given by
where
x is the length (in m) of the element of area (expressed in terms of y)
y is the depth (in m) of the element of area
w is the density of the liquid (in N m-3)
(for water, this is w = 9800 N m-3)
a is the depth at the top of the area in question (in m)
b is the depth at the bottom of the area in question (in m)
Example 1.
Find the force on one side of a cubical container 6.0 cm on an edge if the container is filled with mercury. The weight density of mercury is 133 kN/m3.
- Answer
-
This is the same as having a square plate of sides 6.0 cm submerged in mercury.
This is a very basic example where the width of the plate does not change as we move down the plate.
It is always x = 6.
Also, the depth of the top of the plate is 0, so a = 0.
To apply the formula, we have:
x = 0.06 m
y = is the variable of integration
w = 133 kN m-3 = 133000 N m-3
a = 0
b = 0.06
So we have:
Example 2.
A right triangular plate of base 2.0 m and height 1.0 m is submerged vertically, with the top vertex 3.0 m below the surface.
Find the force on one side of the plate.
- Answer
-
Before we can proceed, we need to find x in terms of y.
Now when x = 0, y = 3 and when x = 2, y = 4.
So we have:
This gives us x = 2y − 6.
To apply the formula, we have:
x = 2y − 6 = 2(y − 3)
y = is the variable of integration
w = 9800 N m-3
a = 3
b = 4
So we have:
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